Question

(i) Write the products formed when benzaldehyde reacts with the following reagents :

(I) 2,4-Dinitrophenylhydrazine

(II) Acetophenone in the presence of dilute NaOH followed by heating

(ii) Give reasons for the following :

(I) Benzoic acid does not undergo Friedel-Crafts reaction.

(II) Carboxylic acids have higher boiling point than alcohols of comparable molecular mass.

(iii) Write simple chemical test to distinguish between propanal and propanone.

Hint:

Benzaldehyde forms a hydrazone with 2,4-DNP and chalcone with acetophenone. Explain the reasons using deactivation and hydrogen bonding. Tollens test or iodoform test separates the pair.

Answer 1

Step 1: Understanding the Question:
Part (i) asks for two products of benzaldehyde. Part (ii) asks for two reasons. Part (iii) asks for one test that separates an aldehyde from a ketone.

Step 2: Part (i)(I), Benzaldehyde with 2,4-DNP:
2,4-Dinitrophenylhydrazine reacts with the carbonyl group of an aldehyde or ketone. Water is lost and a hydrazone forms.
\[ \mathrm{C_6H_5CHO + H_2N-NH-C_6H_3(NO_2)_2 \rightarrow C_6H_5CH=N-NH-C_6H_3(NO_2)_2 + H_2O} \]
The product is benzaldehyde 2,4-dinitrophenylhydrazone. It is a yellow or orange crystalline solid.

Step 3: Part (i)(II), Benzaldehyde with acetophenone and dilute NaOH:
Benzaldehyde has no alpha hydrogen. Acetophenone has alpha hydrogens, so it forms an enolate with dilute NaOH.
The enolate adds to benzaldehyde in a cross aldol reaction. On heating, water is lost from the aldol and a conjugated enone forms. This is the Claisen-Schmidt condensation.
\[ \mathrm{C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{dil.\ NaOH,\ \Delta} C_6H_5CH=CH-CO-C_6H_5 + H_2O} \]
The product is benzalacetophenone (chalcone), also named 1,3-diphenylprop-2-en-1-one.

Step 4: Part (ii)(I), Benzoic acid and Friedel-Crafts reaction:
The \( \mathrm{-COOH} \) group is electron withdrawing, so it deactivates the benzene ring toward electrophilic attack.
Also the catalyst \( \mathrm{AlCl_3} \) is a Lewis acid. It bonds to the lone pair on the oxygen of the carboxyl group and forms a salt. This deactivates the ring further.
So benzoic acid does not undergo the Friedel-Crafts reaction.

Step 5: Part (ii)(II), Boiling point of carboxylic acids:
Both alcohols and acids form hydrogen bonds. But a carboxylic acid has a more polar O-H bond and also a C=O group.
Two molecules join through two hydrogen bonds and form a stable cyclic dimer. Even in the vapour phase, many acids stay as dimers.
An alcohol of similar molar mass forms weaker and fewer hydrogen bonds. So the acid needs more energy to boil and has the higher boiling point.

Step 6: Part (iii), Test for propanal and propanone:
Use Tollens test. Warm each compound with Tollens reagent, which is ammoniacal silver nitrate.
Propanal is an aldehyde, so it reduces \( \mathrm{Ag^+} \) to silver. A shining silver mirror forms on the tube.
\[ \mathrm{CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag + 4NH_3 + 2H_2O} \]
Propanone is a ketone, so it gives no silver mirror.

Final Answer:
(i)(I) Benzaldehyde 2,4-dinitrophenylhydrazone. (i)(II) Benzalacetophenone (chalcone), \( \mathrm{C_6H_5CH=CHCOC_6H_5} \).
(ii)(I) The ring is deactivated by \( \mathrm{-COOH} \), and \( \mathrm{AlCl_3} \) bonds to the carboxyl group. (ii)(II) Acids form strong cyclic hydrogen bonded dimers.
(iii) Propanal gives a silver mirror with Tollens reagent. Propanone does not.
\[ \boxed{\text{Tollens test: propanal gives silver mirror, propanone gives none}} \]